Fields

**topsquark** · February 14th 2006, 06:57 PM

Got another one for y'all. I got part of the way through this, but I'm stuck for a way to continue.

Here's the question:

"Let K be a field, and let R be a subring of K. Suppose that every element of K satisfies a polynomial whose coefficients are in R and whose highest degree coefficient is 1. Show that R is actually a subfield of K."

The question goes on to give a couple of hints:
1) Show that R is an integral domain
2) Given a nonzero $a \epsilon R$ we know that 1/a satisfies a suitable polynomial with coefficients in R. Manipulate this polynomial.

Here's what I've got so far:

Proof: R is an integral domain.
Assume R is a subring of the field K. Then R is an integral domain because any element of R is also an element of K, thus the subring R is commutative and contains no zero-divisors. (End of proof.)

For the second part, here's as far as I've gotten.

By hypothesis, $\forall a\epsilon K$ $\exists \{r_i\} \subset R$ and $n \epsilon Z_{+}$ such that:
$a^n+r_{n-1}a^{n-1}+...+r_0=0$

Now, since $a \epsilon K$ and K is a field $\exists 1/a \epsilon K$ . We may multiply the above equation by $(1/a)^n$ :
$1+r_{n-1}(1/a)+...+r_0(1/a)^n=0$
or
$r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$ .

Again, since $a \epsilon K$ and K is a field $\exists 1/a \epsilon K$ , by hypothesis:
$(1/a)^m+q_{m-1}(1/a)^{m-1}+...+q_0=0$ ; $\{q_i\} \subset R$
where m does not need to be equal to n. We may multiply this equation by $r_0$ :
$r_0(1/a)^m+r_0q_{m-1}(1/a)^{m-1}+...+r_0q_0=0$ .
Compare this with
$r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$ .

I stated before that m was not necessarily equal to n, but these equations together show that the set of all possible values for m (given this a) contains the value n. So set m=n. Furthermore, R contains no zero-divisors, so the second equation is not trivial (that is, all of the coefficients are non-zero.) Finally, the leading coefficient is the same in both equations. Thus I conclude the polynomials are equal in the sense that ALL the coefficients of like terms are equal.

In particular, comparison of the constant term shows $r_0q_0=1$ where both $r_0,q_0 \epsilon R$ . Thus, by definition, $q_0$ is the multiplicative inverse of $r_0$ .

This is as far as I've gotten. I can show by this method that some elements of R have inverses in R, but not all of them. Only those $r_0$ values that come up in polynomials for given values of a will have inverses, not the whole set R.

Any thoughts to finish this, or am I beating down a dead end?

Thanks!
-Dan

**topsquark** · February 16th 2006, 11:49 AM

Originally Posted by topsquark

Got another one for y'all. I got part of the way through this, but I'm stuck for a way to continue.

Here's the question:

"Let K be a field, and let R be a subring of K. Suppose that every element of K satisfies a polynomial whose coefficients are in R and whose highest degree coefficient is 1. Show that R is actually a subfield of K."

The question goes on to give a couple of hints:
1) Show that R is an integral domain
2) Given a nonzero $a \epsilon R$ we know that 1/a satisfies a suitable polynomial with coefficients in R. Manipulate this polynomial.

Here's what I've got so far:

Proof: R is an integral domain.
Assume R is a subring of the field K. Then R is an integral domain because any element of R is also an element of K, thus the subring R is commutative and contains no zero-divisors. (End of proof.)

For the second part, here's as far as I've gotten.

By hypothesis, $\forall a\epsilon K$ $\exists \{r_i\} \subset R$ and $n \epsilon Z_{+}$ such that:
$a^n+r_{n-1}a^{n-1}+...+r_0=0$

Now, since $a \epsilon K$ and K is a field $\exists 1/a \epsilon K$ . We may multiply the above equation by $(1/a)^n$ :
$1+r_{n-1}(1/a)+...+r_0(1/a)^n=0$
or
$r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$ .

Again, since $a \epsilon K$ and K is a field $\exists 1/a \epsilon K$ , by hypothesis:
$(1/a)^m+q_{m-1}(1/a)^{m-1}+...+q_0=0$ ; $\{q_i\} \subset R$
where m does not need to be equal to n. We may multiply this equation by $r_0$ :
$r_0(1/a)^m+r_0q_{m-1}(1/a)^{m-1}+...+r_0q_0=0$ .
Compare this with
$r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$ .

I stated before that m was not necessarily equal to n, but these equations together show that the set of all possible values for m (given this a) contains the value n. So set m=n. Furthermore, R contains no zero-divisors, so the second equation is not trivial (that is, all of the coefficients are non-zero.) Finally, the leading coefficient is the same in both equations. Thus I conclude the polynomials are equal in the sense that ALL the coefficients of like terms are equal.

In particular, comparison of the constant term shows $r_0q_0=1$ where both $r_0,q_0 \epsilon R$ . Thus, by definition, $q_0$ is the multiplicative inverse of $r_0$ .

This is as far as I've gotten. I can show by this method that some elements of R have inverses in R, but not all of them. Only those $r_0$ values that come up in polynomials for given values of a will have inverses, not the whole set R.

Any thoughts to finish this, or am I beating down a dead end?

Thanks!
-Dan

I was finally able to figure it out. I'll post it for the sake of completeness.

Our story so far...

The subring R of the field K is actually an integral domain, so R contains no zero-divisors. Additionally, the hypothesis assures us that $\forall a \in K$ we have that $\exists \{r_i \} \subset R$ and $n \in \mathbb{Z}_{+}$ such that:
$a^n+r_{n-1}a^{n-1}+...+r_0=0$ .
This leads to the result that $r_0$ has a multiplicative inverse in R.

The problem at this point is that we need to prove that an inverse exists for all elements of R. I was finally able to get around that problem:

Since $a^n+r_{n-1}a^{n-1}+...+r_0=0$ , take any $b \in R$ and form the product polynomial:
$(a+b)(a^n+r_{n-1}a^{n-1}+...+r_0)=0$
This produces a new polynomial:
$a^{n+1}+...+br_0=0$ .
By the construction in the previous post (and as repeated above) the existence of this polynomial implies that the element $br_0$ has an inverse in R also. Recall that multiplication in R is commutative (since it is a subring of the field K.) Thus:
$\frac{1}{br_0} \in R$ $\Rightarrow$ $r_0 \frac{1}{br_0} \in R$ $\Rightarrow$ $\frac{1}{b} \in R$ $\forall b \in R$ , since multiplication in R is closed by definition.

So all elements of R have a multiplicative inverse, making the integral domain R a field.

-Dan

**rgep** · February 16th 2006, 01:28 PM

I think you can make it rather shorter. For $r \in R$ , the inverse $r^{-1} \in K$ satisfies a monic equation $r^{-n} + a_{n-1}r^{-n+1} + \cdots + a_1 r^{-1} + a_0 = 0$ with all the $a_i \in R$ . Multiply by $r^{n-1}$ to get $r^{-1} = -a_{n-1} - \cdots - a_0 r^{n-1}$ and the RHS is in $R$ since it is a ring.

**topsquark** · February 16th 2006, 04:36 PM

Originally Posted by rgep

I think you can make it rather shorter. For $r \in R$ , the inverse $r^{-1} \in K$ satisfies a monic equation $r^{-n} + a_{n-1}r^{-n+1} + \cdots + a_1 r^{-1} + a_0 = 0$ with all the $a_i \in R$ . Multiply by $r^{n-1}$ to get $r^{-1} = -a_{n-1} - \cdots - a_0 r^{n-1}$ and the RHS is in $R$ since it is a ring.

Bugger.

I hate it when I miss the easy way...

Thanks for the tip!

-Dan

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